Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is the fair price of this game?

Strategy
Since the expected value of a single roll is 3.5, I will stop on any roll if I get a 4, 5, or 6.
If Fair Price = Expected Value
And Expected Value = P(i_th roll) x E(i_th roll) i=1 to 3
Then Fair Value = P(1 roll) E(1st roll) + P(2 rolls) E(2nd roll) + P(3 rolls) * E(3rd roll)
E(i_th roll) = 3.5 for any roll
P(only 1 roll) = 1/2 :: must roll 4|5|6
P(only 2 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 4|5|6
P(3 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 1|2|3
$\frac{1}{2} * 3.5 + \frac{1}{4} * 3.5 + \frac{1}{4} * 3.5 = $
$\frac{1}{2}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} = $
$\frac{7}{4} + \frac{7}{8} + \frac{7}{8} = $
$\frac{14}{8} + \frac{7}{8} + \frac{7}{8} = $
$\frac{28}{8} = 3.50$
Whn trying to solve this before, I think I made a mistake and end up with something like $4\frac{5}{8} = 4.625$
Should 3.5 be the answer, or should the 1st and 2nd rolls be "wieghted" by what values you must roll to move on (1,2 or 3) instead of the general 3.5 expected value for a single roll?
 A: If you only have one throw left, then the expected amount you will have after the throw, if you do use it, is $\ \frac{7}{2}\ $. Therefore, if your previous throw was $3$ or less, you should use your last throw, but if it was $4$ or more you should not use your last throw, but accept the value of the previous one.
Thus, when you have two throws left, your expected winnings, if you decide to take the next throw, will be
$$
\frac{1}{2}\times\frac{7}{2}+\frac{4+5+6}{6}=\frac{17}{4}\ .
$$
Thus, if your first throw was four or less you should throw again, but if it was $5$ or $6$ you should not throw again, but accept the value your first throw. Your expected winnings, and the fair price of the game, is therefore
$$
\frac{2}{3}\times\frac{17}{4}+\frac{5+6}{6}=\frac{14}{3}\ .
$$
A: There is $\frac{1}{2}$ chance that we get $4, 5$ or $6$ in the first throw and $\frac{1}{2}$ chance that we get $1, 2$ or $3$.
Thus for exactly one  throw , the expected value is  $$\frac{1}{2} \times \frac{4+5+6}{3}=\frac{1}{2} \times 5$$
For exactly two throws , we must get $ \{1, 2  \text {    }\mathrm{or} \text {    }3 \}$ in the first throw and $ \{4, 5  \text {    }\mathrm{or} \text {    }6 \}$ in the $2$nd throw.
The expected value for exactly $2$ throws  is: $$\frac{1}{2} \times 5$$
For exactly 3 throws , we must get $ \{1, 2  \text {    }\mathrm{or} \text {    }3 \}$ in the first $2$ throws and accept whatever we get in the $3$rd throws.
So the expected value for exactly $3$ throws  is $$\frac{1}{4} \times 3.5$$
Thus the final expected value is
$$\frac{1}{2} \times 5 + \frac{1}{4} \times 5 + \frac{1}{4} \times 3.5 = 4.625$$
A: The value of a single roll is $3.5$
With two rolls, keep the first if it is more than $3.5$, so the value is $$\frac12\times5+\frac12\times3.5=4.25$$
With three rolls, keep the first if it is more than $4.25$
